Question

The annual demand for a product is 16,400 units. The weekly demand is 315 units with a standard deviation of 90 units. The cost to place an order is $31.00, and the time from ordering to receipt is four weeks. The annual inventory carrying cost is $0.20 per unit.
a. Find the reorder point necessary to provide a 95 percent service probability.
b. Suppose the production manager is asked to reduce the safety stock of this item by 55 percent. If she does so, what will the new service probability be?

Answer 1

Answer:

a) The reorder point necessary to provide a 95 percent service probability is 1557 units.

b) The Z value of 0.74 corresponds to 77% service probability.

Step-by-step explanation:  

Average weekly demand (d) = 315 units  

The standard deviation of weekly demand (\sigmad) = 90 units  

Lead time (L) = 4 weeks  

At 95% service level value of Z = 1.65  

Reorder point = d x L + safety stock  

$= d \times L + (Z \times \sigma d \times \sqrt L)\\\\= 315 x 4 + (1.65 x 90 x \sqrt 4)\\\\= 1260 +(1.65 x 90 x 2)\\\\= 1260 + 297\\\\= 1557 units$

b) Earlier the safety stock was 297 units(calculated in part a)

Now the safety stock is reduced to 55%.so,55% of 297 = 163.35 units  

So the new safety stock = 297 - 163.35 = 133.65  

$Safety stock = Z \times \sigma d \times \sqrt L\\133.65 = Z x 90 x 2\\133.65 = 180Z\\ Z = 133.65/180\\Z = 0.74$  

The Z value of 0.74 corresponds to 77% service probability.